Sometimes the third condition is stated as f (a) = f (b) = 0,but for the proof, it doesn't matter.

In pictures, we're saying suppose f is a nice smooth function with the same starting and ending height:

If f increases or decreases from its starting height, it needs to turn around and come back in order to end at the same height it started at:

Since f is a nice smooth differentiable function, its derivative at that turn-around point must be 0:

If f doesn't go up or down from its starting point, then f is constant:

In this case, f ' (c) is 0 for every value of c in the interval (a, b).

Rolle's Theorem is reminiscent of the Intermediate Value Theorem. Rolle's Theorem says if f satisfies some assumptions (more mathematically known as hypotheses) then f ' will be zero at some point in (a, b). We could have a constant function, in which case f ' will be 0 infinitely many times:

We could have a function that turns around once:

Or we could have a function that turns around many times:

Rolle's Theorem doesn't tell us where or how many times f' will be zero; it tells us f ' must be zero at least once if the hypotheses are all satisfied.

Sample Problem

Suppose f is not continuous on [a, b]. Then there doesn't need to be any c in (a, b) with f ' (c) = 0. Here's an example:

This function is not continuous. At the point of discontinuity, f ' doesn't exist. At all other points in the interval, f ' is positive:

There is no point c in (a, b) where f ' (c) = 0.

We found earlier that the derivative of the absolute value function doesn't exist at 0. When x is negative the slope of the absolute value function is -1; when x is positive the slope of the absolute value function is -1:

There is no value of c anywhere, in any interval (a, b), with f ' (c) = 0. The derivative of the absolute value function isn't 0 anywhere.

This function doesn't have a derivative of 0 anywhere between a and b.

If a function fails any of the hypotheses, we aren't allowed to use Rolle's Theorem.

Example 1

We need to check that f satisfies all the hypotheses of Rolle's Theorem.

f is a polynomial, so f is continuous on [-2,2].

f is differentiable on (-2,2), since we found that f'(x) = 2x.

f(-2) = 4 and f(2) = 4, so f(-2) = f(2).

Since f satisfies all the hypotheses of Rolle's Theorem, Rolle's Theorem says there must be some c in (-2,2) for which f'(c) = 0.

In this case, f(x) = x2has a "turn-around point" at x = 0, so f'(0) = 0. We can see this from looking at the graph or from finding f'(0), but not from Rolle's Theorem. Rolle's Theorem doesn't tell us where f' is zero, but that it is somewhere.

Exercise 1

Let f(x) = sin(x). This function is differentiable everywhere. Prove that there is some c in (0, 2π) with f'(c) = 0. By looking at the graph of f, determine how many such values of c there are in (0,2π).